Chapter 3: Problem 27
Factor the polynomial completely and find all its zeros. State the multiplicity of each zero. $$P(x)=x^{4}+3 x^{2}-4$$
Short Answer
Expert verified
Zeros are \( 1, -1, \pm 2i \), each with multiplicity 1.
Step by step solution
01
Identify the Quadratic Form
Notice that the polynomial \( P(x) = x^4 + 3x^2 - 4 \) resembles a quadratic in form if we let \( u = x^2 \). Thus, the polynomial becomes \( u^2 + 3u - 4 \).
02
Factor the Quadratic Polynomial
Factor the polynomial \( u^2 + 3u - 4 \). We need two numbers that multiply to \(-4\) and add to \(3\). These numbers are \(4\) and \(-1\). The polynomial can thus be factored as \((u + 4)(u - 1)\).
03
Substitute Back for \(x\)
Substitute \( u = x^2 \) back into the factors obtained: \( (x^2 + 4)(x^2 - 1) \).
04
Factor Further
Recognize that \( x^2 - 1 \) is a difference of squares and can be further factored as \((x - 1)(x + 1)\). Thus, the entire expression becomes \((x^2 + 4)(x - 1)(x + 1)\).
05
Find the Zeros of Each Factor
Set each factor equal to zero and solve: 1. \( x^2 + 4 = 0 \) gives \( x = \pm 2i \).2. \( x - 1 = 0 \) gives \( x = 1 \).3. \( x + 1 = 0 \) gives \( x = -1 \).
06
Determine Multiplicities of Zeros
From the factors, we have the zeros: - \( x = \pm 2i \) each with multiplicity 1 (from \( x^2 + 4 \)).- \( x = 1 \) with multiplicity 1 (from \( x - 1 \)).- \( x = -1 \) with multiplicity 1 (from \( x + 1 \)).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Form
When tackling polynomials, recognizing a structure similar to a quadratic can simplify the factoring process. A quadratic form in its simplest terms is any expression or equation resembling the typical quadratic polynomial, which is \[ ax^2 + bx + c \].For the given polynomial \( P(x) = x^4 + 3x^2 - 4 \), noticing the quadratic form relies on the substitution \( u = x^2 \). By substituting, the equation morphs into \( u^2 + 3u - 4 \), which strongly resembles a familiar quadratic equation.Breaking down this step:
- Identify expressions containing even powers of a variable.
- Replace \( x^{2n} \) terms with a new variable like \( u \) to form a quadratic equation.
Difference of Squares
The difference of squares is a crucial concept when factoring polynomials. It refers to expressions of the pattern \( a^2 - b^2 \), which can be factored into \((a - b)(a + b)\).In our problem, once we substitute back and obtain \((x^2 + 4)(x^2 - 1)\), observe that \(x^2 - 1\) is a perfect example of the difference of squares.To break it down further:
- Recognize the pattern \( a^2 - b^2 \).
- In \( x^2 - 1 \), consider it as \( (x)^2 - (1)^2 \).
- Factor it as \((x - 1)(x + 1)\).
Complex Numbers
In solving polynomials, sometimes we encounter factors that yield complex numbers as solutions. This usually happens when a quadratic does not have real solutions, often indicated by a negative under the square root sign in the quadratic formula.In the factor \( x^2 + 4 \), setting it equal to zero results in the equation \( x^2 = -4 \). Solving this results in \( x = \pm 2i \), where \( i \) is the imaginary unit defined as \( \sqrt{-1} \).To summarize:
- When the discriminant (\( b^2 - 4ac \)) of a quadratic is negative, expect complex solutions.
- If an expression equals \( -b^2 \), the square root yields \( bi \).
- Complex roots often appear in conjugate pairs such as \( 2i \) and \( -2i \). These roots are crucial when completely solving quadratics within polynomial expressions.